integral of e to the power -x is -e to the power -x
Add 1 to the coefficient and divide by that term. int[f(X) by the power rule is this.......X(n + 1)/nint(6X)= 6/2(X2)= 3X2 + C--------------
1.001
The integral of x cos(x) dx is cos(x) + x sin(x) + C
The integral of 2-x = 2x - (1/2)x2 + C.
integral of e to the power -x is -e to the power -x
∫ xn dx = xn+1/(n+1) + C (n ≠-1) C is the constant of integration.
∫ f(x)nf'(x) dx = f(x)n + 1/(n + 1) + C n ≠-1 C is the constant of integration.
What do you mean? As this is a calculus question, I presume that you are asking for a derivative or integral The derivative of any function of the form ƒ(x) = a * x ^ n is ƒ'(x) = a * n * x ^ (n-1) The integral of any function of the form ∫ a*x ^ n is a / (n+1) * x ^ (n+1) + C Your function that you gave is 1 / x^(2) which is equal to: x^(-2) Thus the derivative is: -2 * x^(-3) And the integral is: -x^(-1) + C
If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.
You add one to the power, and then divide by the power that it has so you would have: Integral of x = (x^2)/2 Integral of x^2 = (x^3)/3 Etc.
The antiderivative, or indefinite integral, of ex, is ex + C.
∫ f'(x)(af(x) + b)n dx = (af(x) + b)n + 1/[a(n + 1)] + C C is the constant of integration.
∫ f(x)/[f(x) + g(x)]n dx = ∫ 1/[f(x) + g(x)]n - 1 dx - ∫ g(x)/[f(x) + g(x)]n dx
The integral of a single term of a polynomial, in the form of AxN is (A/N+1) x (N+1). The first integral of 2x is x2 + C. The second integral of 2x is the first integral of x2 + C, which is 1/3x3 + Cx + C.
Add 1 to the coefficient and divide by that term. int[f(X) by the power rule is this.......X(n + 1)/nint(6X)= 6/2(X2)= 3X2 + C--------------
Using information from the Wolframalpha site. It seems that this integral can't be expressed as a finite amount of standard functions; you can go to the Wolfram Alpha site, and type "integral x^x", to get a series expansion if you are interested.